Problem: You have found the following ages (in years) of all 6 gorillas at your local zoo: $ 20,\enspace 13,\enspace 8,\enspace 12,\enspace 6,\enspace 3$ What is the average age of the gorillas at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Solution: Because we have data for all 6 gorillas at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $6$ ages and divide by $6$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\mu} = \dfrac{20 + 13 + 8 + 12 + 6 + 3}{{6}} = {10.3\text{ years old}} $ Find the squared deviations from the mean for each gorilla. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $20$ years $9.7$ years $94.09$ years $^2$ $13$ years $2.7$ years $7.29$ years $^2$ $8$ years $-2.3$ years $5.29$ years $^2$ $12$ years $1.7$ years $2.89$ years $^2$ $6$ years $-4.3$ years $18.49$ years $^2$ $3$ years $-7.3$ years $53.29$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{94.09} + {7.29} + {5.29} + {2.89} + {18.49} + {53.29}} {{6}} $ $ {\sigma^2} = \dfrac{{181.34}}{{6}} = {30.22\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{30.22\text{ years}^2}} = {5.5\text{ years}} $ The average gorilla at the zoo is 10.3 years old. There is a standard deviation of 5.5 years.